Book XI. solid CV, asa the solid CD to the solid CV. But as the so. un lid AB to the solid CV, sob is the base EH to the base 27.5. 15. NP; for the solids AB, CV are of the same altitude; and 532. 11. 35. u. as the solid CD to CV, soc is the base MP to the base d 1. 6. PT, and sod is the straight line MC to CT; and CT is equal to AG. Therefore, as the base EHto the base NP, so is MC to AG. Wherefore the bases of the solid parallelopi. peds AB, CD are reciprocally proportional to their alti-... tudes. Let now the bases of the solid parallelopipeds AB, CD be reciprocally proportional to their altitudes, viz, as the base Eh to the base NP, so the altitude of the solid CD to the altitude of the so- K B R D lid AB; the solid AB GAF IM NX is equal to the solid CD. Let the insisting lines be, as before, at right angles to the he bases. Then, if the base EH be equal to A N the base NP, since EH is to NP, as the altitude of the so lid CD is to the altitude of the solid AB, therefore the alti• A. 5. tude of CD is equal e to the altitude of AB. But solid pa rallelopipeds upon equal bases, and of the same altitude, ; f31. 11. are equal f to one another: therefore the solid AB is equal to the solid CD. . But let the bases EH, NP be unequal, and let EH be R D CN, MP to the same altituto the solid as solide CV: And therefore as the solid AB to the solid CV, Book XL so is the solid CD to the solid CV; that is, e:ch of the solids AB, CD has the same ratio to the solid CV; and © 25. 11. therefore the solid AB is equal to the solid CD. Second general case. Let the insisting straight lines FE, BL, GA, KH; XN, DO, MC, RP not be at right angles to the bases of the solids; and from the points F, B, K, G; X, D, R, M draw perpendiculars to the planes in which are the bases EH, NP, meeting those planes in the points S, Y, V, T; Q, I, U, Z; and complete the solids FV, XU, which are parallelopipeds, as was proved in the last part of Prop. 31 of this Book. In this case, likewise, if the solids AB, CD be equal, their bases are reciprocally proportional to their altitudes, viz. the base EH to the base NP, as the altitude of the solid CD to the altitude of the solid AB. Because the solid AB is equal to the solid CD, and that the solid BT is equals to the solid BA, 5 29. or 30. 11. for they are upon the same base FK, and of the same alti tude; and that the solid DC is equals to the solid DZ, being upon the same base XR, and of the same altitude; therefore the solid BT is equal to the solid DZ: But the bases are reciprocally proportional to the altitudes of equal solid parallelopipeds of which the insisting straight lines are at right angles to their b:ises, as before was proved : Therefore as the base FK to the base XR, so is the altitude of the solid DZ to the altitude of the solid BT: And the base FK is equal to the base EH, and the base XR to the base NP: Wherefore, as the base EH to the base NP, so is the altitude of the solid DZ to the altitude of the solid BT: But the altitudes of the solids DZ, DC, as also of the solids BT, BA are the same. Therefore as the base EA to the base NP, so is the altitude of the solid CD to the alti Book XI. tude of the solid AB ; that is, the bases of the solid paral- altitudes. A of AB : But the altitudes of the solids AB, BT are the same, as also of CD and DZ: therefore as the base FK to the base XR, so is the altitude of the solid DZ to the altitude of the solid BT: Wherefore the bases of the solids BT, DZ are reciprocally proportional to their altitudes; and their insisting straight lines are at right angles to the bases; wherefore, as was before proved, the solid BT is 829. or 30. equal to the solid DZ: But BT is equals to the solid BA, and DZ to the solid DC, because they are upon the same bases, and of the same altitude. Therefore the solid AB is equal to the solid CD. Q. E. D. ll. Book XI, PROP. XXXV. THEOR. If, from the vertices of two equal plane angles, See N. there be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the sides of those triangles, each to each; and if in the lines above the planes there be taken any points, and from them perpendiculars be drawn to the planes in which the first-pamed angles are: And from the points in which they meet the planes, straight lines be drawn to the vertices of the angles first named; these straight lines shall contain equal angles with the straight lines which are above the planes of the angles. Let BAC, EDF be two equal plane angles: and from the points A, D let the straight lines AG, DM be elevated above the planes of the angles making equal angles with their sides, each to each, viz. the angle GAB equal to the angle MDE, and GAC to MDF; and in AG, DM let any points G, M be taken, and from them let perpendiculars GL, MN be drawn to the planes BAC, EDF, meeting M these planes in the points L, N; and join LA, ND: The angle GAL is equal to the angle MDN. Make AH equal to DM, and through H draw HK parallel to GL: But GL is perpendicular to the plane BAC; wherefore HK is perpendiculara to the same plane: From' 8. 11. the points K, N, to the straight lines AB, AC, DE, DF, draw perpendiculars KB, KC, NE, NF; and join HB, BC, ME, Book XI. EF: Because HK is perpendicular to the plane BAC, the plane HBK which passes through HK is at right anglesb 618. 11. to the plane BAC; and AB is drawn in the plane BAC at right angles to the common section BK of the two planes; , therefore AB is perpendicularc to the plane HBK, and * 4 Def. 11. az Def. 11. makes right angles d with every straight line meeting it in that plane : But BH meets it in that plane; therefore ABH is a right angle: For the same reason DEM is a right angle, and is therefore equal to the angle ABH. And the angle HAB is equal to the angle MDE. Therefore in the two triangles HAB, MDE there are two angles in one equal to two angles in the other, each to each, and one side equal to one side, opposite to one of the equal an gles in each, viz. HA equal to DM; therefore the remain* 26. 1. ing sides are equale, each to each: Wherefore AB is equal to DE. In the same manner, if HC and MF be joined, it and DF; and the angle BAC is equal to the angle EDF; 14. 1. wherefore the base BC is equalf to the base EF, and the remaining angles to the remaining angles: The angle ABC is therefore equal to the angle DEF: And the right angle ABK is equal to the right angle DEN, whence the remaining angle CBK is equal to the remaining angle FEN: For the same reason, the angle BCK is equal to the angle EFN: Therefore in the two triangles BCK, EFN, there are two angles in one equal to two angles in the other, each to each, and one side equal to one side ad jacent to the equal angles in each, viz. BC equal to EF; the other sides, therefore, are equal to the other sides; BK then is equal to EN: and AB is equal to DE: wherefore AB, BK are equal to DE, EN; and they contain right angles; wherefore the base AK is equal to the base DN: And since |